Find the moment of inertia of a sphere about an axis through its center. The radius of the sphere is a, and the homogeneous density is e.

Question

Find the moment of inertia of a sphere about an axis through its center. The radius of the sphere is a, and the homogeneous density is [latex]\varrho[/latex].

Accepted Answer

We use cylindrical coordinates [latex](r,\varphi, z)[/latex]. The z-axis is the rotation axis. For the corresponding moment of inertia, we have

[latex] \Theta= \varrho \int\limits_{sphere}{r^{2} dV}.[/latex]

The center of the sphere is at z = 0. The equation for the spherical surface then reads

[latex]x^{2} +y^{2} + z^{2} = a^{2} or r^{2} +z^{2} = a^{2}.[/latex]

We write out the integration limits:

[latex] \Theta= \varrho \int\limits_{0}^{2π}{d\varphi} \int\limits_{−a}^{a}{dz} \int\limits_{0}^{\sqrt{a^{2}−z^{2}}}{r^{3} dr}[/latex]

or

[latex] \Theta=2π \varrho \int\limits_{−a}^{a}{\left[\frac{1}{4} r^{4}\right]^{\sqrt{a^{2}−z^{2}}}_{0} dz} = \frac{π}{2} \varrho \int\limits_{−a}^{a}{(a^{2} −z^{2})^{2}dz}.[/latex]

Integration over z yields

[latex] \Theta= πa^{5} \varrho \frac{8}{15}= \frac{4}{3}πa^{3} \varrho \frac{2}{5}a^{2}.[/latex]

Since the total mass of the sphere is given by [latex]M = (4/3)πa^{3}\varrho[/latex], it follows that

[latex] \Theta= \frac{2}{5} Ma^{2}[/latex].

Question 11.3: Find the moment of inertia of a sphere about an axis through...

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